Probability

Question 1

Complementary Events

Answer 1

P(A) + P(Not A) = 1

When two events are complementary, knowing the probability that one event will occur allows us to calculate the probability that the complement will occur.

Question 2

Probability of A and B:

Independent Events

Answer 2

P(A and B) = P(A) x P(B)

• When two events, A and B, are independent, the probability of event A and event B occurring
• Successive occurrences of an event are each independent of each other
• Independent events and their complements: If 2 events, A and B, are independent, then their complements are also independent.
  → (not A) and (not B) are independent
  → (not A) and B are independent
  → A and (not B) are independent

Question 3

Probability of A and B:

Dependent Events

Answer 3

P(A and B) = P(A) x P(B | A)

• When two events, A and B, are dependent, the probability of event A and event B occurring
• P(B | A) = probability that B will occur given that A has already occurred

Question 4

Mutually Exclusive Event Probability

Answer 4

• When A and B are mutually exclusive: P(A or B) = P(A) + P(B)
• When A and B are NOT mutually exclusive: P(A or B) = P(A) + P(B) - P(A and B)

Question 5

Multiple Outcomes

Answer 5

• When an event has more than one possible outcome, each possible outcome must be considered when calculating the probability that the event will occur.
• To determine the actual probability: (number of outcomes producing the event) x (probability of one outcome)
• Leverage the permutation formula for indistinguishable items

Example: The probability that it rains in Town X on any given day is 40%. What is the probability that it will rain in Town X on exactly 3 days in a certain 4-day period?

• Probability of raining on 3 out of 4 days: (2/5)³ x (3/5)
• # of permutations for raining 3 out of 4 days: XXXY → 4! / 3! = 4
• Multiply the two together: 96/625

Question 6

Combinatorics + Probability

Answer 6

When you’re presented with a dependent probability problem with multiple outcomes, consider using the combination method in which you separately calculate the numerator and denominator of:

Probability of an event = (number of favorable outcomes) / (total number of outcomes)

Question 7

Probability with Permutations: Example #1

Answer 7

A computer algorithm will randomly create an eight-digit number from the digits 5, 5, 5, 5, 6, 6, 6, and 7. What is the probability that the number created is 66,655,557?

Answer: 1 / 280
→ Total = 8! / (3! x 4!) = 280
→ That number is 1 of the ways. So 1 / 280

Question 8

Baye’s Theorem

Answer 8

P(A | B) = P(B | A) x P(A) / P(B)

Question 9

Probability: Hard Question Example

Answer 9

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

Total # of ways of assigning 4 letters to 4 envelopes is 4!=24.

Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct).

ABCD(envelopes)
ACDB(letters)
ADBC(letters)

(When A is in the right envelope other three have only 2 possible incorrect arrangements)

As we have 4 letters, total # of ways 4*2=8